Forces and Free-Body Diagrams — Practice Problem Solutions

1. Magnitude $=\sqrt{6^2+8^2}=\sqrt{100}=10\text{ N}$.

2. A single dot with two arrows: one pointing straight up labeled $F_N$ (normal force), one pointing straight down labeled $F_g$ (gravity/weight), drawn the same length since the book is motionless (equilibrium).

3. Magnitude $=\sqrt{5^2+12^2}=\sqrt{169}=13\text{ N}$; direction $=\arctan\left(\dfrac{12}{5}\right)\approx67.4°$ above horizontal.

4. $mg\sin25°$.

5. Invalid reasoning. A genuine Newton's-third-law pair acts on two different objects — one force on each. Here, both the tension and the sign's weight act on the same object (the sign), so they form an equilibrium pair (the balance required by Newton's first/second law), not a third-law pair. The tension's actual third-law partner is the sign pulling down on the rope; gravity's third-law partner is the sign pulling up on the Earth.

6. a) Three forces on the crate: gravity straight down, the normal force perpendicular to the ramp's surface, and friction pointing up the ramp (opposing the crate's tendency to slide down). b) The component of gravity along the ramp is $mg\sin20°$. Friction must match this in magnitude because the crate is motionless — for the net force along the ramp to be zero (equilibrium), the up-ramp friction force has to exactly balance the down-ramp pull of gravity.

7. a) Yes — before the rope is cut, the FBD shows tension (up) and gravity (down) with equal magnitudes, since the sign is motionless; the net force is zero. b) No — the instant the rope is cut, the tension force disappears from the FBD entirely (it's not weakened, it's gone), leaving only gravity acting on the sign. The net force is now unbalanced and downward, so the sign begins to accelerate — it's in free fall.

8. Magnitude $=\sqrt{9^2+12^2}=\sqrt{225}=15\text{ N}$ — (B).